LeetCode Populating Next Right Pointers in Each Node

这题很有新意，刚开始想到后序遍历，可是对于给的例子5和6之间怎么连接呢？无论怎么遍历5和6之间都隔好几个数，递归引用传参数的方法肯定不行，题中又说常量的空间，对于完全二叉树最多也不过64层吧，所以每一层设定一个指针用于建立连接。传递指针数组不太容易，用了全局变量解决。如下：

// LeetCode_PopulatingNextRightPointers.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <iostream>using namespace std;struct TreeLinkNode {  int val;  TreeLinkNode *left, *right, *next;  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} };TreeLinkNode* layers[64];void preOrderLayer(TreeLinkNode *root,int layer)//,TreeLinkNode *&layers{if (layers[layer]==NULL){layers[layer] = root;}else{layers[layer]->next = root;layers[layer] = root;}if(root->left)preOrderLayer(root->left,layer+1);if(root->right)preOrderLayer(root->right,layer+1);}void connect(TreeLinkNode *root) {if(root==NULL)return ;for (int i=0;i<64;i++)layers[i] = NULL;int layer = 0;preOrderLayer(root,layer);for (int i=0;i<64&&layers[i]!=NULL;i++){layers[i]->next = NULL;}}int _tmain(int argc, _TCHAR* argv[]){TreeLinkNode *pa = new TreeLinkNode(1);  TreeLinkNode *pb = new TreeLinkNode(2);  TreeLinkNode *pc = new TreeLinkNode(3);  TreeLinkNode *pd = new TreeLinkNode(4);//TreeLinkNode *pd = NULL;//  TreeLinkNode *pe = new TreeLinkNode(5);  TreeLinkNode *pf = new TreeLinkNode(6);  TreeLinkNode *pg = new TreeLinkNode(7);  pa->left = pb;  pa->right = pc;  pb->left = pd;  pb->right = pe;  pc->left = pf;  pc->right = pg; connect(pa);system("pause");return 0;}